Question

We can extend the technique of trigonometric substitution to the hyperbolic functions. Use Theorem 2.20 and the identity ${\cosh }^{2}x-{\sinh }^{2}x=1$to solve each integral in Exercises 87– 90 with an appropriate hyperbolic substitution $x=a\sinh u\text{or}x=a\cosh u.$. (These exercises involve hyperbolic functions.)

$\int \frac{1}{\sqrt{x^2+4}}dx$

Short answer

The value of the integral is${\sinh }^{-1}\left(\frac{x}{2}\right)+C$.

Step-by-step solution

Step 1. Given information.

The given integral is$\int \frac{1}{\sqrt{x^2+4}}dx$

Step 2. Substitution.

To solve the integral,

Let$$\begin{gathered} x=\sinh \left(u\right) \\ dx=\cosh \left(u\right)du \end{gathered}$$

${\sinh }^2\left(u\right)+1={\cosh }^2\left(u\right)$.

Now,

$$\begin{gathered} \frac{1}{\sqrt{x^2+4}}=\frac{1}{\sqrt{4{\sinh }^2\left(u\right)+4}} \\ =\frac{1}{2\sqrt{{\sinh }^2\left(u\right)+1}} \\ =\frac{1}{2\sqrt{{\cosh }^2\left(u\right)}} \\ =\frac{1}{2\cosh \left(u\right)} \end{gathered}$$

Step 3. Value of the integral.

Now, we can write,

$$\begin{gathered} \int \frac{1}{\sqrt{x^2+4}}dx \\ =\int 1du \\ =u \\ ={\sinh }^{-1}\left(\frac{x}{2}\right)+C \end{gathered}$$